2018 AMC 12B Problems/Problem 22

Problem

Consider polynomials $P(x)$ of degree at most $3$ , each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ . How many such polynomials satisfy $P(-1) = -9$ ?

$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$

Solution 1

Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $P(-1)=-a+b-c+d=-9.$ Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.

We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or $a'+b+c'+d=9.$ By the stars and bars argument, there are $\binom{9+4-1}{4-1}=\boxed{\textbf{(D) } 220}$ ordered quadruples $(a',b,c',d).$

~pieater314159 ~MRENTHUSIASM

Solution 2

Suppose our polynomial is equal to $ax^3+bx^2+cx+d$ Then we are given that $9=b+d-a-c.$ Then the polynomials $cx^3+bx^2+ax+d$ , $ax^3+dx^2+cx+b,$ $cx^3+dx^2+ax+b$ also have $b+d-a-c=-9$ when $x=-1.$ So the number of solutions must be divisible by 4. So the answer must be $\boxed{\textbf{D}.}$

Solution 3

As before, $-9=-A+B-C+D$ . This is $9=(A+B)-(C+D)$ . Rephrased, how many two sums of integers from 0-9 have a difference of 9. Make a chart of pairs between these two sets: ${0,1,2,3,4,5,6,7,8,9}$ ${9,10,11,12,13,14,15,16,17,18}$ Observe how there is one way to sum 2 numbers to 0 and two ways to 1, 3 ways to 2, and so on. At 9, there are 10 possible ways. Recall that only integers between 0-9 are valid. Now observe how there is 1 way to to sum to 18 in this fashion (9+9), 2 ways to sum to 17, and so forth again (to optionally prove that this pattern holds, apply stars and bars up to 9 and notice the symmetry).

The answer then is the number of ways to write each component of each pair. This is $(1*10)+(2*9)+(3*8) ...$ or, since it's symmetrical between sum of 4 and 5, $[2\sum\limits_{i=1}^{5} i(11-i)]$ . Use summation rules to finally get $\boxed{\textbf{D)}220}$ .

~BJHHar

2018 AMC 12B (Problems • Answer Key • Resources)
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2018 AMC 12B Problems/Problem 22

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also